Counting Odd-Only Subsets with One Element from a Prefix
Combinatorics · Medium · Free problem
Let $[n] = \{1, 2, \ldots, n\}$ and fix
\le k \le n$. Count the number of subsets $S \subseteq [n]$ satisfying both: 1. $S$ contains exactly one element from $\{1, 2, \ldots, k\}$. 2. Every element of $S$ is odd.
Solve for the general case and evaluate at $n = 8$, $k = 3$.
Hints
Even numbers are excluded entirely -- restrict attention to odd integers from the start.
Count the odds in $\{1,\ldots,k\}$ (there are $\lceil k/2 \rceil$) and the odds in $\{k+1,\ldots,n\}$ (there are $\lceil n/2 \rceil - \lceil k/2 \rceil$).
Choose exactly one element from the first pool and any subset from the second pool; multiply the counts: $\lceil k/2 \rceil \cdot 2^{\lceil n/2 \rceil - \lceil k/2 \rceil}$.
Worked Solution
How to Think About It: You need exactly one odd from $\{1, \ldots, k\}$ and any collection of odds from $\{k+1, \ldots, n\}$. Even numbers are completely excluded. So count the odds in each segment, choose one from the first and any subset from the second, and multiply by the multiplication principle.
Quick Estimate: For $n=8$, $k=3$: odds in $\{1,2,3\}$ are $\{1,3\}$, so 2 choices for the required element. Odds in $\{4,5,6,7,8\}$ are $\{5,7\}$, so
^2 = 4$ subsets from that pool. Total:
\times 4 = 8$.
Approach: Count odd integers in each segment using the ceiling formula, then apply the multiplication principle.
Formal Solution:
Number of odd integers in $[r]$: The positive integers alternate odd/even starting with 1 (odd). So in $\{1, \ldots, r\}$, the count of odd integers is $\lceil r/2 \rceil$. Verification: $r=1 \to 1$, $r=2 \to 1$, $r=3 \to 2$, $r=4 \to 2$, etc.
Building the subset:
Step 1: Choose exactly one odd from $\{1, \ldots, k\}$. There are $\lceil k/2 \rceil$ odd numbers in this range. So there are $\lceil k/2 \rceil$ choices.
Step 2: Choose any subset of odds from $\{k+1, \ldots, n\}$. The number of odds in $[n]$ is $\lceil n/2 \rceil$; of these, $\lceil k/2 \rceil$ are in $[k]$ and are excluded from this step. So the remaining odd count is $\lceil n/2 \rceil - \lceil k/2 \rceil$, and the number of subsets is
^{\lceil n/2 \rceil - \lceil k/2 \rceil}$.
By the multiplication principle: $\text{Answer} = \lceil k/2 \rceil \cdot 2^{\lceil n/2 \rceil - \lceil k/2 \rceil}.$
Evaluation at $n=8$, $k=3$: - $\lceil 3/2 \rceil = 2$ (odds in $\{1,2,3\}$: namely 1 and 3) - $\lceil 8/2 \rceil = 4$ (odds in $\{1,\ldots,8\}$: namely 1, 3, 5, 7) - Odds in $\{4,5,6,7,8\}$: $4 - 2 = 2$ (namely 5 and 7)
This is a straightforward application of the multiplication principle, once you decompose the problem correctly. The two constraints (exactly one from $[k]$, all odd) partition the work into two independent choices: which odd from $[k]$ to include, and which odds from $\{k+1,\ldots,n\}$ to include. Independence of these choices means you multiply, not add.
The ceiling formula $\lceil r/2 \rceil$ for the number of odd integers in $[r]$ is worth internalizing -- it is cleaner than parity-casing and appears in many combinatorics problems involving subsets with parity constraints. A common mistake is to treat $k$ and $n$ as arbitrary without checking the parity formula; verifying on small examples (as done in the quick estimate) prevents errors.